ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Alexx on October 11, 2010, 08:25:39 pm

Title: How would i go about this question>?
Post by: Alexx on October 11, 2010, 08:25:39 pm: Okay, so we've gone through this in class so many times and each time we're getting a different answer.

Does the 3.50A apply to each of the six cells or the whole battery?

Can someone please walk me through it?
Title: Re: How would i go about this question>?
Post by: akira88 on October 11, 2010, 08:37:27 pm: "A particular lead-acid battery delivers a current of 3.50A..."
I would assume it is just the battery as it does not say "each cell delivers a current of 3.50A".
Hope that helps :)
Title: Re: How would i go about this question>?
Post by: physics on October 11, 2010, 08:40:15 pm: apparently its the whole thing
Title: Re: How would i go about this question>?
Post by: matt123 on October 11, 2010, 08:47:14 pm: Quote from: akira88 on October 11, 2010, 08:37:27 pm
"A particular lead-acid battery delivers a current of 3.50A..."
I would assume it is just the battery as it does not say "each cell delivers a current of 3.50A".
Hope that helps :)

yeah agreed.
its the whole thing = 3.50 amp.s

therefore for part I
since n(e) = Q/f

you get 3.5 X 20 X 60 / 96500
which gives you 0.0435 mol of electrons

then since its +2 electrons .. every +2 gives you 1 mol .
hence its 0.0435 / 2 ... = 0.02175 mol of the pb02

hmm part ii)
i would find the mass overall of the battery
then divide it by 6.

some 1 correct me if im wrong

hope that helps
Title: Re: How would i go about this question>?
Post by: Mao on October 11, 2010, 11:10:48 pm: The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0217*6, and the answer to part ii) is 0.0217*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.
Title: Re: How would i go about this question>?
Post by: matt123 on October 12, 2010, 07:53:38 am: Quote from: Mao on October 11, 2010, 11:10:48 pm
The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0435*6, and the answer to part ii) is 0.0435*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.

just out of curiosity.

since there are 2 electrons
why dont u go 0.435 / 2
giving 0.2 abouts?
Title: Re: How would i go about this question>?
Post by: Mao on October 12, 2010, 11:32:48 am: Quote from: matt123 on October 12, 2010, 07:53:38 am
Quote from: Mao on October 11, 2010, 11:10:48 pm
The cells are connected in series. I'm fairly sure the answer to part b i) is 0.0435*6, and the answer to part ii) is 0.0435*mass(PbSO4)

It's been a while since I looked at cells in this much detail, but if I remember correctly (I don't have a schematic handy atm), 3.50A current passes through each of the 6 cells, and in turn produces 12V when one pair of the redox only creates 2V.

just out of curiosity.

since there are 2 electrons
why dont u go 0.435 / 2
giving 0.2 abouts?

Misread your post, apologies. Previous post fixed.
Title: Re: How would i go about this question>?
Post by: crayolé on October 13, 2010, 12:47:49 am: So the current doesn't get split up between the 6 cells?
Title: Re: How would i go about this question>?
Post by: Mao on October 13, 2010, 05:58:41 am: Quote from: crayola on October 13, 2010, 12:47:49 am
So the current doesn't get split up between the 6 cells?

Connected in series --> the same current goes through each cell connected 'head-to-tail', the voltage adds up (each cell is only 2V, a 6-cell battery is 12V).